How to shuffle rows of a matrix, say order 4, with the help of set having 4 elements, several times?

Let

R=[1,2,3,4;5,6,7,8;9,10, 11,12;13,14,15,16];
k=[3,4,1,2];
for i=1:4
R1(i,:)=R(k(i),:);
end
   for i=1:4
    R2(i,:)=R1(k(i),:);
    end
 for i=1:4
    R3(i,:)=R2(k(i),:);
    end
R3

My output will be R3. I want to do this using loop

[M]

You should format your code to make it more readable.

And where are you stuck exactly?

[Abdul Gaffar]

Thanks, but still having problem

[Birdman]

Seen my answer?

Answer 1

[Jan]

The for i loop over k is not needed:

R = [1,2,3,4; 5,6,7,8; 9,10,11,12; 13,14,15,16];
k = [3,4,1,2];
for i = 1:3
  R = R(k, :);
end

Now R is modified inplace.

If you R is huge, it is cheaper to shuffle an index vector instead:

k = [3,4,1,2];
v = 1:4;
for i = 1:3
  v = v(k);
end
Rfinal = R(v, :)

[Abdul Gaffar]

Perhaps, you haven't understand my question. I want to change the rows of matrix with the help of set (sort of permutation ), 3 times

[Jan]

@ABDUL: And this is exactly what my code does. Did you try it?

R = randi([1,10], 4, 4)
k = [3,4,1,2]
R = R(k, :)
R = R(k, :)
R = R(k, :)

Now the rows of R are shuffled according to k 3 times. Because the 1st and 3rd, as well as 2nd an 4th rows are swapped, the sequence of R is repeating. But for longer k there will be more states.

The same in a loop:

for j = 1:3
  R = R(k, :)
end

The line

R = R(k, :)

is just a simpler version of

 for i = 1:4
   R(i, :) = R(k(i), :);
 end

As far as I understand your question, you do not need R1 and R2, so I've overwritten R.

Applying the permutation to the index vector is more efficient, because you do not access the complete matrix R in each iteration and the result is the same.

What is the different to what you want? Do you want to change the index vector k in each iteration?

[Abdul Gaffar]

Thanks